probability question from Sklansky book

[Scooterdoo]

I was showing a friend who is new to poker Sklansky's Hold'em Poker beginner book and specifically going over the probability section in the back. One of David's probabilities didn't seem right to me, or maybe I wasn't reading it correctly. He says on page 108:

"In a ten-handed game the chance that someone holds both an ace and another card of a specified suite is about 9 percent..."

I read this to mean that if I pick any suit, there is a 9% chance that someone holds Ax of that suit; the implications are that if there are 3 of the same suit on the board than 9 percent of the time in a ten handed game someone has the nut flush. This seems way too high. I can perhaps see that there is a 9% probability that someone has Axs, but not Ax of a specific suit. BTW, he does say that if you have two of the specific suit for a flush the probability goes down to about 6% that someone else has the A-nut flush.

If David is correct can someone please explain the math on this?

[Nottom]

[ QUOTE ]
I read this to mean that if I pick any suit, there is a 9% chance that someone holds Ax of that suit; the implications are that if there are 3 of the same suit on the board than 9 percent of the time in a ten handed game someone has the nut flush. This seems way too high. I can perhaps see that there is a 9% probability that someone has Axs, but not Ax of a specific suit. BTW, he does say that if you have two of the specific suit for a flush the probability goes down to about 6% that someone else has the A-nut flush.

[/ QUOTE ]

Without going into the math, this assumption is incorrect. David is saying that without knowing anything about the cards, there is a 9% chance someone has a suited Ace. If you have a mono-tone flop then it would greatly reduce the chance that someone has a suited Ace in that suit. If the flop is monotone and you have a flush then this chance would be reduced even more.

[rdu $teve]

I dont know the odds of flopping the nut flush. However the odds of flopping ANY flush are 117 to 1, when you hold to suited cards. Maybe that can get you somewhere.

[aloiz]

He's correct if there was no other information specified. At a 10 person table preflop the odds that a player will be deal Axs in a given suit = 10*12/combin(52,2) =~.0905.

aloiz

[Scooterdoo]

I guess the problem that is meaniful is this. Assuming on the board there are 3 to a flush and you've made the flush -- to simplify it let's assume that you have a king high flush -- what is the probability that someone has the A high flush? Using David's example it would be around 6%. Even this seems high to me.

[aloiz]

If you have a king high flush with three to the flush on the final board, the odds that someone has an A high flush against n players = n * C(8,2)/C(45,2) =~ .028n


aloiz

[Murph]

[ QUOTE ]
If you have a king high flush with three to the flush on the final board, the odds that someone has an A high flush against n players = n * C(8,2)/C(45,2) =~ .028n

[/ QUOTE ]

Hi aloiz,

I want to learn more about such calculations etc. Unfortunately, I have only a basic math background. Could you please restate your calculation in plain English (i.e. explain the abbreviations and the underlying rationale for same). I hope to then be able to use your explanation to better understand future posts.

Are there some good basic books on probability and figuring odds that you would recommend?

Thanks.

[teddyboy]

Sklansky's "getting the best of it" is a good book on calcualtions for Poker (Bayes Theorem is key)

[Murph]

Thanks. Just ordered it.

[Ryendal]

If you have a king high flush with three to the flush on the final board, the odds that someone has an A high flush against n players = n * C(8,2)/C(45,2) =~ .028n


Don't you think it's ---> n * 8 /C(45,2) ?

[aloiz]

Yea I was saying one thing and thinking another. Should be n * 7/C(45,2) as there is only 7 ways to make an Axs combination with 5 cards of that suit already gone.

aloiz

[belowpoverty]

NM